如何从文件中提取路径

Question

我需要读取一个文件，其中包含其他文件的路径，有类型和其他数据。文件看起来像这样，

LIST OF SUB DIRECTORIES:
Advanced System Optimizer 3
ashar wedding and home pics
components
Documents and Settings
khurram bhai
media
new songs
Office10
Osama
Program Files
RECYCLER
res
Stationery
System Volume Information
Templates
WINDOWS

LIST OF FILES:
.docx  74421
b.com  135168
ChromeSetup.exe  567648
Full & final.CPP  25884
hgfhfh.jpg  8837
hiberfil.sys  267964416
myfile.txt.txt  0
pagefile.sys  402653184
Shortcut to 3? Floppy (A).lnk  129
Thumbs.db  9216
vcsetup.exe  2728440
wlsetup-web.exe  1247056

我只需要提取出文件的路径名称，并将它们保存在一个数组中，但我坚持它。这是我的代码，

// read a file into memory
#include <iostream>
#include <fstream>
using namespace std;
int main () {
  int length;
  char str[600];
  ifstream is;
  is.open ("test.txt", ios::binary );
  // get length of file:
  is.seekg (0, ios::end);
  length = is.tellg();
  is.seekg (0, ios::beg);

  // read data as a block:
  is.read (str,length);
  //**find the path of txt files in the file and save it in an array...Stuck here**
  is.close();
  return 0;
}

我不知道下一步该做什么。即使我使用strstr()来查找。txt，无论它何时出现，我如何获得它的整个路径?

Answer 1

也许你应该看看boost文件系统库。

它提供了你需要的东西。

这应该是一个如何工作的例子。虽然我没有试过，但它应该可以编译。

boost::filesystem::path p("test.txt");
boost::filesystem::path absolutePath = boost::filesystem::system_complete(p);
boost::filesystem::path workDir = absolutePath.parent_path();
std::vector<std::string> file;
std::string line;
std::ifstream infile ("test.txt", std::ios_base::in);
while (getline(infile, line, 'n'))
{
    file.push_back (line.substr(0, line.find_first_of(" ")));
}
std::vector<std::wstring> fullFileNames;
for(std::vector<std::string>::const_iterator iter = file.begin(); iter != file.end(); ++iter)
{
    boost::filesystem::path newpath= workDir / boost::filesystem::path(*iter);
    if(!boost::filesystem::is_directory(newpath) && boost::filesystem::exists(newpath))
    {
        fullFileNames.push_back(newpath.native().c_str());
    }
}

当然，它缺少各种错误检查

Answer 2

如果您只需要提取路径，并且文件总是看起来像这样，您可以逐行读取文件并使用string::find查找空格的第一次出现，并为每个条目创建一个子字符串。

size_t index = str.find(" ");
if(index != string::npos) // sanity checing
{
   string path = str.substr(0, index);
   //do whatever you want to do with the file path
}

Answer 3

您想要完成的实际上是演示如何在cplusplus.com上使用string::find_last_of的示例代码:

void SplitFilename (const string& str)
{
  size_t found;
  cout << "Splitting: " << str << endl;
  found=str.find_last_of("/\");
  cout << " folder: " << str.substr(0,found) << endl;
  cout << " file: " << str.substr(found+1) << endl;
}

Answer 4

如果您想获得当前目录中给定文件的整个路径，下面的代码可以帮您完成，当然使用boost:

#include <iostream>
#define BOOST_FILESYSTEM_VERSION 3
#include <boost/filesystem.hpp>
namespace fs = boost::filesystem;
int main()
{
  fs::path my_path("test.txt");
  if(fs::is_regular_file(my_path)) // Sanity check: the file exists and is a file (not a directory)
  {
    fs::path wholePath = fs::absolute(my_path);
    std::cout << wholePath.string() << std::endl;
  }
  return 0;
}